Strictly increasing solutions of non-autonomous
difference equations arising in hydrodynamics
Lukáš Rachůnek and Irena Rachůnková
Department of Mathematics, Faculty of Science, Palacký University,
tř. 17. listopadu 12, 77146 Olomouc, Czech Republic,
e-mail: [email protected]
Abstract. The paper provides conditions sufficient for the existence of strictly
increasing solutions of the second-order non-autonomous difference equation
n
x(n + 1) = x(n) +
n+1
2 x(n) − x(n − 1) + h2 f (x(n)) ,
n ∈ N,
where h > 0 is a parameter and f is Lipschitz continuous and has three real zeros
L0 < 0 < L.
In particular we prove that for each sufficiently small h > 0 there exists
∞
a solution {x(n)}∞
n=0 such that {x(n)}n=1 is increasing, x(0) = x(1) ∈ (L0 , 0)
and limn→∞ x(n) > L. The problem is motivated by some models arising in
hydrodynamics.
Keywords. Non-autonomous second-order difference equation, strictly increasing solutions, discretization.
Mathematics Subject Classification 2000. 39A11, 39A12, 39A70
1
Formulation of problem
We will investigate the following second-order non-autonomous difference equation
n
x(n + 1) = x(n) +
n+1
2 x(n) − x(n − 1) + h2 f (x(n)) ,
n ∈ N,
(1.1)
where f is supposed to fulfil
L0 < 0 < L,
f ∈ Liploc [L0 , ∞),
xf (x) < 0 for x ∈ (L0 , L) \ {0},
1
f (L0 ) = f (0) = f (L) = 0,
(1.2)
f (x) ≥ 0 for x ∈ (L, ∞),
(1.3)
∃B̄ ∈ (L0 , 0) such that
Z L
f (z) dz = 0.
(1.4)
B̄
Let us note that f ∈ Liploc [L0 , ∞) means that for each [L0 , A] ⊂ [L0 , ∞) there
exists KA > 0 such that |f (x) − f (y)| ≤ KA |x − y| for all x, y ∈ [L0 , A]. A simple
example of a function f satisfying (1.2)–(1.4) is f (x) = c(x − L0 )x(x − L), where
c is a positive constant.
A sequence {x(n)}∞
n=0 which satisfies (1.1) is called a solution of equation
(1.1). For each values B, B1 ∈ [L0 , ∞) there exists a unique solution {x(n)}∞
n=0
of equation (1.1) satisfying the initial conditions
x(0) = B,
x(1) = B1 .
(1.5)
Then {x(n)}∞
n=0 is called a solution of problem (1.1), (1.5).
In [17] we have shown that equation (1.1) is a discretization of differential
equations which generalize some models arising in hydrodynamics or in the nonlinear field theory, see [5], [6], [8], [12]. Increasing solutions of (1.1), (1.5) with
B = B1 ∈ (L0 , 0) have a fundamental role in these models. Therefore, in [17], we
have described the set of all solutions of problem (1.1), (1.6), where
x(0) = B,
x(1) = B,
B ∈ (L0 , 0).
(1.6)
In this paper, using [17], we will prove that for each sufficiently small h > 0 there
exists at least one B ∈ (L0 , 0) such that the corresponding solution of problem
(1.1), (1.6) fulfils
x(0) = x(1),
lim x(n) > L,
n→∞
{x(n)}∞
n=1 is increasing.
(1.7)
Note that an autonomous case of (1.1) was studied in [16]. We would like to
point out that recently there has been a huge interest in studying the existence
of monotonous and nontrivial solutions of nonlinear difference equations. For
papers during last three years see for example [1], [2], [4], [9]–[11], [13]–[15], [19],
[20]–[24]. A lot of other interesting references can be found therein.
2
Four types of solutions
Here we present some results of [17] which we need in next sections. In particular,
we will use the following definitions and lemmas.
Definition 2.1 Let {x(n)}∞
n=0 be a solution of problem (1.1), (1.6) such that
{x(n)}∞
n=1 is increasing,
Then {x(n)}∞
n=0 is called a damped solution.
2
lim x(n) = 0.
n→∞
(2.1)
Definition 2.2 Let {x(n)}∞
n=0 be a solution of problem (1.1), (1.6) which fulfils
{x(n)}∞
n=1 is increasing,
lim x(n) = L.
n→∞
(2.2)
Then {x(n)}∞
n=0 is called a homoclinic solution.
Definition 2.3 Let {x(n)}∞
n=0 be a solution of problem (1.1), (1.6). Assume that
there exists b ∈ N, such that {x(n)}b+1
n=1 is increasing and
x(b) ≤ L < x(b + 1).
(2.3)
Then {x(n)}∞
n=0 is called an escape solution.
Definition 2.4 Let {x(n)}∞
n=0 be a solution of problem (1.1), (1.6). Assume that
there exists b ∈ N, b > 1, such that {x(n)}bn=1 is increasing and
0 < x(b) < L,
x(b + 1) ≤ x(b).
(2.4)
Then {x(n)}∞
n=0 is called a non-monotonous solution.
Lemma 2.5 [17] (On four types of solutions)
∞
Let {x(n)}∞
n=0 be a solution of problem (1.1), (1.6). Then {x(n)}n=0 is just one
of the following four types:
(I)
(II)
{x(n)}∞
n=0 is an escape solution;
{x(n)}∞
n=0 is a homoclinic solution;
(III)
{x(n)}∞
n=0 is a damped solution;
(IV)
{x(n)}∞
n=0 is a non-monotonous solution.
Lemma 2.6 [17] (Estimates of solutions)
Let {x(n)}∞
n=0 be a solution of problem (1.1), (1.6). Then there exists a maximal
b ∈ N ∪ {∞} satisfying
x(n) ∈ [B, L)
for n = 1, . . . , b,
x(n) ∈ [B, L) for n ∈ N,
if b ∈ N,
if b = ∞.
(2.5)
Further, if b > 1, then moreover
{x(n)}bn=1
is increasing,
q
∆x(n) < h (L − 2L0 )M0 + h2 M0
(2.6)
(2.7)
for n = 1, . . . , b − 1 if b ∈ N, and for n ∈ N if b = ∞, where
M0 = max{|f (x)|: x ∈ [L0 , L]}.
3
(2.8)
In [17] we have proved that the set consisting of damped and non-monotonous
solutions of problem (1.1), (1.6) is nonempty for each sufficiently small h > 0.
This is contained in the next lemma.
Lemma 2.7 [17] (On the existence of non-monotonous or damped solutions)
Let B ∈ (B̄, 0), where B̄ is defined by (1.4). There exists hB > 0 such that if
h ∈ (0, hB ], then the corresponding solution {x(n)}∞
n=0 of problem (1.1), (1.6) is
non-monotonous or damped.
In Section 4 of this paper we prove that also the set of escape solutions of
problem (1.1), (1.6) is nonempty for each sufficiently small h > 0. Note that in
our next paper [18] we prove this assertion for the set of homoclinic solutions.
3
Properties of solutions
Now, we provide other properties of solutions important in the investigation of
escape solutions.
Lemma 3.1 Let {x(n)}∞
n=0 be an escape solution of problem (1.1), (1.6). Then
∞
{x(n)}n=1 is increasing.
Proof. Due to (1.1), {x(n)}∞
n=0 fulfils
∆x(n) =
n
n+1
2 ∆x(n − 1) + h2 f (x(n)) ,
n ∈ N.
(3.1)
According to Definition 2.3 there exists b ∈ N, such that {x(n)}b+1
n=1 is increasing
and (2.3) holds. By (1.3) we get f (x(b + 1)) ≥ 0. Consequently, by (3.1) and
2
∆x(b) > 0 and f (x(b + 2)) ≥ 0. Similarly ∆x(b + j) ≥
(2.3), ∆x(b + 1) ≥ b+1
b+2
2
b+j
b+1+j
∆x(b + j − 1) and
∆x(b + j) ≥
b+1
b+1+j
!2
∆x(b),
j ∈ N.
This yields that {x(n)}∞
n=1 is increasing.
(3.2)
Lemma 3.2 Assume that f (x) = 0 for x > L. Choose an arbitrary % > 0. Let
∞
B1 , B2 ∈ (L0 , 0) and let {x(n)}∞
n=0 and {y(n)}n=0 be a solution of problem (1.1),
(1.6) with B = B1 and B = B2 , respectively. Let KL be the Lipschitz constant
for f on [L0 , L]. Then
2K
|x(n) − y(n)| ≤ |B1 − B2 |e%
∆x(n) − ∆y(n) h
L
,
≤ |B1 − B2 |%KL e%
where n ∈ N, n ≤ h% .
4
(3.3)
2K
L
,
(3.4)
Proof. By (3.1) we have
(j + 1)2 ∆x(j) − j 2 ∆x(j − 1) = h2 j 2 f (x(j)),
j ∈ N.
(3.5)
Summing it for j = 1, . . . , k, we get by (1.6),
∆x(k) = h2
k
X
1
j 2 f (x(j)),
(k + 1)2 j=1
k ∈ N.
(3.6)
Summing it again for k = 1, . . . , n − 1, we get
x(n) = B1 + h2
k
X
1
j 2 f (x(j)),
2
(k
+
1)
j=1
k=1
n ∈ N,
k
X
1
j 2 f (y(j)),
2
(k
+
1)
j=1
k=1
n ∈ N.
n−1
X
and similarly
y(n) = B2 + h2
n−1
X
From this and by using summation by parts we easily obtain
k
X
1
j 2 |f (x(j)) − f (y(j))|
|x(n) − y(n)| ≤ |B1 − B2 | + h
2
k=1 (k + 1) j=1
2
n−1
X
≤ |B1 − B2 | + (n − 1)h2 KL
n−1
X
|x(j) − y(j)|,
n ∈ N.
j=1
By the discrete analogue of the Gronwall-Bellman inequality (see e.g. [7], Lemma
4.34), we get
|x(n) − y(n)| ≤ |B1 − B2 |e(n−1)
2 h2 K
L
for n ∈ N,
which yields (3.3).
By (3.6) and (3.3) we have for n ∈ N, n ≤ h% ,
∆x(n) − ∆y(n) h
≤ hKL
n
X
≤h
n
X
1
j 2 |f (x(j)) − f (y(j))|
(n + 1)2 j=1
2K
|x(j) − y(j)| ≤ |B1 − B2 |%KL e%
L
.
j=1
5
4
Existence of escape solutions
∞
Lemma 4.1 Assume that C ∈ (L0 , B̄) and {Bk }∞
k=1 ⊂ (L0 , C). Let {xk (n)}n=0
be a solution of problem (1.1), (1.6) with B = Bk , k ∈ N. For k ∈ N choose a
maximal bk ∈ N ∪ {∞} such that xk (n) ∈ [Bk , L) for n = 1, . . . , bk if bk is finite,
k
and for n ∈ N if bk = ∞, and {xk (n)}bn=1
is increasing if bk > 1. Then there
∗
∗
exists h > 0 such that for any h ∈ (0, h ], there exists a unique γk ∈ N, γk < bk ,
such that
xk (γk ) ≥ C, xk (γk − 1) < C.
(4.1)
Moreover, if the sequence {γk }∞
k=1 is unbounded, then there exists ` ∈ N such
∞
that the solution {x` (n)}n=0 of problem (1.1), (1.6) with B = B` ∈ (L0 , B̄) is an
escape solution.
Proof. Choose h0 > 0 such that
q
h0 (L − 2L0 )M0 + h20 M0 < |C|.
(4.2)
For k ∈ N denote by {xk (n)}∞
n=0 a solution of problem (1.1), (1.6) with B = Bk .
The existence of bk is guaranteed by Lemma 2.6. By Lemma 2.5, {xk (n)}∞
n=0
is just one of the types (I)–(IV), and if h ∈ (0, h0 ], then the monotonicity of
k
{xk (n)}bn=0
yields a unique γk ∈ N, γk < bk , satisfying (4.1).
For h ∈ (0, h0 ), consider the sequence {γk }∞
k=1 and assume that it is unbounded. Then we have
lim γk = ∞.
(4.3)
k→∞
(Otherwise we take a subsequence.) Assume on the contrary that for any k ∈ N,
∞
{xk (n)}∞
n=0 is not an escape solution. Choose k ∈ N. If {xk (n)}n=0 is damped,
then by Definition 2.1, we have bk = ∞ and
xk (bk ) := lim xk (n) = 0,
k→∞
∆xk (bk ) := lim ∆xk (n) = 0.
k→∞
(4.4)
If {xk (n)}∞
n=0 is homoclinic, then by Definition 2.2, we have bk = ∞ and
xk (bk ) := lim xk (n) = L,
k→∞
∆xk (bk ) := lim ∆xk (n) = 0.
k→∞
(4.5)
If {xk (n)}∞
n=0 is non-monotonous, then by Definition 2.4, we have bk < ∞ and
xk (bk ) ∈ (0, L),
∆xk (bk ) ≤ 0.
(4.6)
To summarize if {xk (n)}∞
n=0 is not an escape solution, then by (4.4), (4.5) and
(4.6), we have
xk (bk ) ∈ [0, L], ∆xk (bk ) ≤ 0.
(4.7)
Since ∆xk (0) = 0, there exists γ̄k ∈ N satisfying
γk ≤ γ̄k < bk ,
∆xk (γ̄k ) = max{∆xk (j): γk ≤ j ≤ bk − 1}.
6
(4.8)
Consider (3.5) with x = xk . By dividing it by j 2 , multiplying such obtained
equality by xk (j + 1) − xk (j − 1) and summing in j from 1 to n we get
(∆xk (n))2 − h2
n
X
f (xk (j))(xk (j + 1) − xk (j − 1))
j=1
n
X
2j + 1
=−
∆xk (j)(xk (j + 1) − xk (j − 1)),
j2
j=1
(4.9)
n ∈ N.
Denote
Ek (n + 1) = (∆xk (n))2 − h2
n
X
f (xk (j))(xk (j + 1) − xk (j − 1)).
(4.10)
j=1
Then we get
Ek (n + 1) = −
n
X
2j + 1
∆xk (j)(xk (j + 1) − xk (j − 1)),
j2
j=1
n ∈ N.
(4.11)
Let us put n = γk − 1 and n = bk − 1 to (4.11) and subtract. By (4.7) and (4.8)
we get
Ek (γk ) − Ek (bk ) =
bX
k −1
j=γk
2j + 1
∆xk (j)(xk (j + 1) − xk (j − 1))
j2
(4.12)
2γk + 1
≤2
∆xk (γ̄k )(L − L0 ).
γk2
Let us put n = γk − 1 and n = bk − 1 to (4.10) and subtract. We get
Ek (γk ) − Ek (bk ) = (∆xk (γk − 1))2 − (∆xk (bk − 1))2
2
+2h
bX
k −1
f (xk (j))
j=γk
xk (j + 1) − xk (j − 1)
.
2
(4.13)
Choose ε > 0 and h1 > 0 such that
ε<
1Z L
f (z) dz,
2 C
h1 M0 <
√
ε.
(4.14)
Let bk < ∞. Then (4.6) holds. Since ∆xk (bk − 1) > 0, f (xk (bk )) < 0 and
∆xk (bk ) ≤ 0, (3.1) yields
bk + 1
bk
!2
|∆xk (bk )| + ∆xk (bk − 1) = h2 |f (xk (bk ))|,
and hence
√
0 < ∆xk (bk − 1) ≤ −h2 f (xk (bk )) < h2 M0 < h ε for h ∈ (0, h1 ].
7
(4.15)
Clearly, if bk = ∞, then by (4.4) and (4.5), inequality (4.15) holds, as well.
Having in mind (1.2) and (1.3), we deduce similarly as in the proof of Theorem
2.7 that there exists δ > 0 such that if
xk (j + 1) − xk (j − 1)
< δ,
2
then
j = γk , . . . , bk − 1,
bX
k −1
xk (j + 1) − xk (j − 1) Z L
f (xk (j))
>
f (z) dz − ε.
2
C
j=γk
(4.16)
(4.17)
Let h2 > 0 be such that
h2
q
(L − 2L0 )M0 + h2 M0 < δ.
(4.18)
If h ∈ (0, h2 ], then (2.7) implies (4.16) and hence (4.17) holds.
Now, let us put h∗ = min{h0 , h1 , h2 } and choose h ∈ (0, h∗ ]. Then, (4.2),
(4.14), (4.18), (4.13)–(4.17) yield
2
2
Ek (γk ) − Ek (bk ) > −h ε + 2h
2
= 2h
Z L
C
Z L
!
f (z) dz − ε
C
!
(4.19)
3
f (z) dz − ε > h2 ε > 0.
2
Finally, (4.12) and (4.19) imply
0 < h2 ε < Ek (γk ) − Ek (bk ) ≤ 2
and
2γk + 1
∆xk (γ̄k )(L − L0 ),
γk2
γk2
h2 ε
·
< ∆xk (γ̄k ).
2(L − L0 ) 2γk + 1
Letting k → ∞, we obtain by (4.3), that limk→∞ ∆xk (γ̄k ) = ∞, contrary to
(4.16). Therefore an escape solution {x` (n)}∞
n=0 of problem (1.1), (1.6) with
B = B` ∈ (L0 , B̄) must exist.
Now, we are in a position to prove the next main result.
Theorem 4.2 (On the existence of escape solutions)
There exists h∗ > 0 such that for any h ∈ (0, h∗ ] there exists an escape solution
{x` (n)}∞
n=0 of problem (1.1), (1.6) for some B = B` ∈ (L0 , B̄).
Proof. Choose h > 0 C ∈ (L0 , B̄) and let KL be the Lipschitz constant for f
on [L0 , L]. Consider a sequence {Bk }∞
k=1 ⊂ (L0 , C) such that limk→∞ Bk = L0 .
Then, for each m ∈ N there exists km ∈ N such that
|Bkm − L0 | < e−m
8
2K
L
(C − L0 ).
(4.20)
Let x0 (0) = x0 (n) = L0 for n ∈ N. Then the sequence {x0 (n)}∞
n=0 is the unique
solution of problem (1.1), (1.6) with B = L0 . Let {xk (n)}∞
n=0 be a solution of
∞
problem (1.1), (1.6) with B = Bk , k ∈ N, and let {γk }k=1 be the sequence of
Lemma 4.1. Then it suffices to prove that {γk }∞
k=1 is unbounded. According to
Lemma 3.2, for each m ∈ N,
|xkm (n) − x0 (n)| ≤ |Bkm − L0 |em
2K
L
,
n≤
n≤
m
,
h
m
.
h
(4.21)
Consequently, (4.20) and (4.21) give
|xkm (n) − x0 (n)| ≤ C − L0 ,
and hence
xkm (n) ≤ C,
n≤
m
.
h
Therefore
m
, m ∈ N,
h
which yields that {γk }∞
k=1 is unbounded. Hence the assertion follows from Lemma
4.1.
γkm (n) ≥
Acknowledgments
The paper was supported by the Council of Czech Government MSM 6198959214.
The authors thank the referees for valuable comments.
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